# HOW TO DRAW A CONTOUR MAP CALCULUS

I have provided a regimen to check out that it is an ellipse yet I want to understand the procedure of reasoning to actually draw the contour map myself.

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$x^2+y^2+xy=C$ for $C=0,1,2,3,...$

I can not seem to acquire it right into an ellipse develop. What must I do?Thanks!

I assume that you have some knowledge of direct algebra.

The quadratic create $x^2 + 2 frac12 xy + y^2$ deserve to likewise be written as a dot product in between 2 vectors

$$x^2 + 2 frac12 xy + y^2 = langle left (eginarraycc 1 & 1/2 \ 1/2 & 1 endarray ight ) cdot left (eginarrayc x \ y endarray appropriate ) , left (eginarrayc x \ y endarray est ) angle= langle M v, v angle$$

Let"s look a little bit at the matrix of this quadratic create

$$M = left(eginarraycc 1 & 1/2 \ 1/2 & 1 endarray ight )$$

It is a genuine symmetric matrix so it has genuine eigenworths and also it has a orthonormal basis of eigenvectors. The eigenvalues are $3/2$ and $1/2$ and also we likewise obtain unit eigenvectors:

$$left(eginarraycc 1 & 1/2\ 1/2 & 1 endarray ight ) cdot left (eginarrayc 1/sqrt2\ 1/sqrt2endarray appropriate ) = 3/2 cdot left(eginarrayc 1/sqrt2\ 1/sqrt2endarray est )\left(eginarraycc 1 &1/2\1/2 & 1 endarray appropriate ) cdot left (eginarrayc -1/sqrt2\ 1/sqrt2endarray appropriate ) = 1/2 cdot left(eginarrayc -frac1sqrt2\ 1/sqrt2endarray appropriate )$$

Because of this we have the equality:

$$left(eginarraycc 1 & 1/2\ 1/2 & 1 endarray ight ) cdot left(eginarraycc 1/sqrt2 & -1/sqrt2 \ 1/sqrt2 & 1/sqrt2 endarray ight) =left(eginarraycc 1/sqrt2 & -1/sqrt2 \ 1/sqrt2 & 1/sqrt2 endarray ight) cdot left(eginarraycc 3/2 & 0\ 0 & 1/2 endarray ideal )$$or $$left(eginarraycc 1 & 1/2\ 1/2 & 1 endarray ight ) = left(eginarraycc 1/sqrt2 & -1/sqrt2 \ 1/sqrt2 & 1/sqrt2 endarray ight) cdot left(eginarraycc 3/2 & 0\ 0 & 1/2 endarray est ) cdot left(eginarraycc 1/sqrt2 & -1/sqrt2 \ 1/sqrt2 & 1/sqrt2 endarray ight)^-1$$we created the matrix of the quadratic develop $M$ as $$M = Rcdot D cdot R^-1$$

wbelow $D$ is the diagonal matrix $left(eginarraycc 3/2 & 0\ 0 & 1/2 endarray ight )$ developed through the eigenworths and $R$ is the rotation matrix created with the matching unit eigenvectors $R = left(eginarraycc 1/sqrt2 & -1/sqrt2 \ 1/sqrt2 & 1/sqrt2 endarray ight) = left(eginarraycc cosfracpi4 & -sin fracpi4 \ sin fracpi4 & cosfracpi4 endarray ight)$

The equation in the vector $v$ giving the curve is $$langle Mv, v angle = C$$

or $$langle R D R^-1 v, v angle = C$$

The rotation matrix $R$ preserves the dot product $langle cdot, cdot angle$ so we have actually $langle R D R^-1 v, v angle = langle D R^-1 v, R^-1 v angle$

Thus the equation of the curve is $$langle D R^-1 v, R^-1 v angle = C$$

Denote by $w = R^-1 v$. We have $langle D w, w angle = C$ and also this equation in $w$ provide an ellipse in standard create $E_0$. Our curve consists of all $v$ so that $w = R^-1 v in E_0$, that is $v in R E_0$.

Therefore, our curve is the rotation of the ellipse $E_0$ by the matrix $R$. Keep in mind that $R$ is a rotation of angle $fracpi4$ counterclockwise.

$E_0$ is the ellipse $3/2cdot x^2 + 1/2cdot y^2 = C$. Our curve $E$ is the rotation counterclockwise of $E_0$ by angle $fracpi4$

In basic, if the matrix of the quadratic form is $M = R_alpha D R_alpha^-1$, then the curve $$langle Mv, v angle = C$$ is the rotation by angle $alpha$ of the conic$$langle D w, w angle = C$$

Obs: It"s customary that for ellipses in typical create the much longer axis is along the $x$-axis. Because of this, it seems preferable to use the decomposition $M = R D R^-1$ where the diagonal of $D$ has aspects in boosting order. We alert that our curves are likewise rotations through angle $fracpi4$ clockwise of ellipses $1/2cdot x^2 + 3/2cdot y^2 =C$.